Lời giải:

$x^2-x^2y-y+8x+7=0$

$\Leftrightarrow x^2+8x+7=y(x^2+1)$

$\Leftrightarrow y=\frac{x^2+8x+7}{x^2+1}$

$\Leftrightarrow y=\frac{(x^2+1)+8x+6}{x^2+1}=1+\frac{8x+6}{x^2+1}$

Áp dụng bđt AM-GM ta có:
$x^2+\frac{1}{4}\geq |x|\geq x$
$\Rightarrow x^2+1\geq x+\frac{3}{4}=\frac{4x+3}{4}$

$\Rightarrow \frac{8x+6}{x^2+1}\leq \frac{2(4x+3)}{\frac{4x+3}{4}}=8$

$\Rightarrow y\leq 1+8=9$

Vậy $y_{\max}=9$

$x^2=\frac{1}{4}$; $x\geq 0\Rightarrow x=\frac{1}{2}$

pt\(\Leftrightarrow x^2\left(1-y\right)+8x+7-y=0\) (1)

Ta có :\(\Delta\)_(x)=\(-y^2+8y+9\)(do làm biếng  nên làm ra denta luôn)

Để tồn tại MAX y thì PT (1) có ngiệm nên \(\Delta\ge0\) \(\Leftrightarrow-y^2+8y+9\ge0\)

\(\Leftrightarrow-y^2-y+9y+9\ge0\Leftrightarrow-y\left(y+1\right)+9\left(y+1\right)\ge0\)

\(\Leftrightarrow\left(y+1\right)\left(9-y\right)\ge0\)

Giải BPT ta được : \(-1\le y\le9\)

\(\Rightarrow\) Max y =9. Thay y=9 vào (1)\(\Rightarrow x=\dfrac{1}{2}\)

Vậy Max y=9\(\Leftrightarrow x=\dfrac{1}{2}\)

đề bài sai r bn ơi phải là +10 chứ ko phải +8 đâu nhá

\(x^2+2xy+6x+6y+2y^2+8=0\\ \Leftrightarrow\left(x+y\right)^2+6\left(x+y\right)+y^2=-8\)

Ta có \(y^2\ge0\Leftrightarrow\left(x+y\right)^2+6\left(x+y\right)\le-8\)

\(\Leftrightarrow\left(x+y\right)^2+6\left(x+y\right)+9\le1\\ \Leftrightarrow\left(x+y+3\right)^2\le1\\ \Leftrightarrow\left|x+y+3\right|\le1\\ \Leftrightarrow-1\le x+y+3\le1\\ \Leftrightarrow2012\le B\le2014\)

\(B_{min}=2012\Leftrightarrow\left\{{}\begin{matrix}x+y+2016=2012\\y=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-4\\y=0\end{matrix}\right.\)

\(B_{max}=2014\Leftrightarrow\left\{{}\begin{matrix}x+y+2016=2014\\y=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-2\\y=0\end{matrix}\right.\)

\(x^2+2xy+6x+6y+2y^2+8=0\)

\(\Leftrightarrow\left(x^2+2xy+y^2\right)+\left(6x+6y\right)+9+y^2-1=0\)

\(\Leftrightarrow\left(x+y\right)^2+6\left(x+y\right)+9=1-y^2\)

\(\left(x+y+3\right)^2=1-y^2\)

Do \(VP=1-y^2\le1\forall x\) \(\Rightarrow VT=\left(x+y+3\right)^2\le1\)

\(\Leftrightarrow-1\le x+y+3\le1\)

\(\Leftrightarrow-1+2013\le x+y+3+2013\le1+2013\)

\(\Leftrightarrow2012\le x+y+2016\le2014\) hay \(2012\le B\le2014\)

B đạt MIN là 2012 \(\Leftrightarrow\hept{\begin{cases}y=0\\x+y+3=-1\end{cases}\Rightarrow\hept{\begin{cases}y=0\\x=-4\end{cases}}}\)

B đạt MAX là 2014 \(\Leftrightarrow\hept{\begin{cases}y=0\\x+y+3=1\end{cases}\Leftrightarrow\hept{\begin{cases}y=0\\x=-2\end{cases}}}\)

\(x^2+2xy+6x+6y+2y^2+8=0\)

\(\Leftrightarrow\left(x+y\right)^2+6\left(x+y\right)+9=1-y^2\)

\(\Leftrightarrow\left(x+y+3\right)^2=1-y^2\)

Ta thấy : \(1-y^2\le1\forall y\) \(\Rightarrow\left(x+y+3\right)^2\le1\)

\(\Rightarrow-1\le x+y+3\le1\)

\(\Rightarrow-1+2013\le x+y+3+2013\le1+2013\)

\(\Rightarrow2012\le x+y+2016\le2014\)

Vậy ta có : 

+) Min \(B=2012\) . Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}y=0\\x+y+3=-1\end{cases}\Leftrightarrow}\hept{\begin{cases}y=0\\x=-4\end{cases}}\)

+) Max \(M=2014\). Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}y=0\\x+y+3=1\end{cases}\Leftrightarrow}\hept{\begin{cases}y=0\\x=-2\end{cases}}\)

chiu

j

x² + 2y² + 2xy - 6x - 2y + 13 = 0

<=> ( x² + 2xy + y² - 6x - 6y + 9 ) + ( y² + 4y + 4 ) = 0

<=> [ ( x² + 2xy + y² ) - ( 6x + 6y ) + 9 ] + ( y + 2 )² = 0

<=> [ ( x + y )² - 2( x + y ).3 + 3² ] + ( y + 2 )² = 0

<=> ( x + y - 3 )² + ( y + 2 )² = 0

Ta có : \(\hept{\begin{cases}\left(x+y-3\right)^2\\\left(y+2\right)^2\end{cases}}\ge0\forall x,y\Rightarrow\left(x+y-3\right)^2+\left(y+2\right)^2\ge0\forall x,y\)

Dấu "=" xảy ra <=> x = 5 ; y = -2

Thế x = 5 ; y = -2 vào A ta được :

\(A=\frac{5^2-7\cdot5\cdot\left(-2\right)+52}{5-\left(-2\right)}=\frac{25+70+52}{7}=\frac{147}{7}=21\)

HPT\(\Leftrightarrow\hept{\begin{cases}x=3m+2-y\\3\left(3m+2-y\right)-2y+m-11=0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x=3m+2-y\\-5y+10m-5=0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x=3m+2-\left(2m-1\right)\\y=2m-1\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x=m+3\\y=2m-1\end{cases}}\)

Ta co:

\(x^2-y^2=\left(m+3\right)^2-\left(2m-1\right)^2=-3m^2+10m+8=-3\left(m-\frac{5}{3}\right)^2+\frac{49}{3}\le\frac{49}{3}\)

Dau '=' xay ra khi \(m=\frac{5}{3}\)

\(\Rightarrow\left(x;y\right)=\left(\frac{14}{3};\frac{7}{3}\right)\)

Vay cap nghiem (x;y) de \(x^2-y^2\)dat max la \(\left(\frac{14}{3};\frac{7}{3}\right)\)